Question

1. sean works in a warehouse one day, he pushes a cardboard box of inventory across the floor. the box has a mass of 17.0 kg, and the coefficient of kinetic friction between the cardboard and the floor is 0.22. when sean releases the box, what is the boxs rate of acceleration?

$overrightarrow{a}=-2.16m/s^{2}$
$overrightarrow{a}=-36.7m/s^{2}$
$overrightarrow{a}=3.74m/s^{2}$
$overrightarrow{a}=5.22m/s^{2}$

Explanation:

Step1: Calculate the frictional force

The frictional force \(F_f=\mu_k N\). Since \(N = mg\) (normal force equals weight when on a horizontal surface), \(F_f=\mu_kmg\). Here, \(\mu_k = 0.22\), \(m = 17.0\space kg\), and \(g=9.8\space m/s^2\). So \(F_f=0.22\times17.0\times9.8\).

$$F_f = 0.22\times17\times9.8= 36.652\space N$$

Step2: Apply Newton's second law \(F = ma\)

When Sean releases the box, the only horizontal force acting on it is the frictional force \(F_f\). According to \(F = ma\), \(a=\frac{F}{m}\). Since the frictional force opposes the motion, \(F=-F_f\). So \(a=\frac{-F_f}{m}\). Substitute \(F_f = 36.652\space N\) and \(m = 17.0\space kg\) into the formula:

$$a=\frac{- 36.652}{17}=- 2.156\approx - 2.16\space m/s^2$$

Answer:

\(\vec{a}=-2.16\space m/s^2\) (the first option)